Can I know the Fourier coefficients of a function that uniformly converges? if so why is it?

Question

For example the function

$$f(x) = \sum\frac{1}{2^n}\sin(2^nx)$$ we can see that the function uniformly converges

Can I know that the Fourier coefficients are $$\hat{f}(x)=\frac{1}{2^n}$$

if so why is that possible?

Answer 1

Denote $s_n(x) = \frac{1}{2^n}\sin(2^nx)$. $f$ is the continuous uniform limit of the series $\sum s_n$.

Therefore the Fourier coefficients of $f$ exist as $\int f(x) \sin nx, \int f(x) \cos nx$ exist and $a_n = 0$, $b_{2^p} = \frac{1}{2^p}$ and $b_n = 0$ for $n \neq 2^p$ for all $n \in \mathbb N$. This is by definition of those coefficients.

And $f$ is the sum of its Fourier sum also due to its definition:

$$f(x) = \sum_{n=0}^\infty\frac{1}{2^n}\sin(2^nx)$$

Answer 2

$$f(x) \sim \sum_k b_k \sin k x$$ is a Fourier sine series of an odd function with period $2\pi.$

We have $$f(x) \sim \sum_n \frac{1}{2^n} \sin 2^n x,$$

so

$$\bbox[5px, border: 1.25pt solid green]{b_k = \left\{ \begin{aligned} \frac{1}{k}, &\quad \text{when }\exists n \text{ s.t. }k=2^n,\\ 0, &\quad \text{otherwise.} \end{aligned}\right.}$$

$$f(x) \sim \sin x + \frac{1}{2} \sin 2x + \frac{1}{4} \sin 4x + \frac{1}{8} \sin 8x + \cdots$$

We could also check that $b_k$ arises from the formula

$$b_k = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin k x \, dx.$$

This checks because the sinusoids are orthogonal:

$$ \frac{1}{\pi}\int_{-\pi}^\pi \sin k x \sin l x \, dx = \delta_{kl}.$$